{ "data": { "question": { "questionId": "1661", "questionFrontendId": "1557", "categoryTitle": "Algorithms", "boundTopicId": 381525, "title": "Minimum Number of Vertices to Reach All Nodes", "titleSlug": "minimum-number-of-vertices-to-reach-all-nodes", "content": "

Given a directed acyclic graph, with n vertices numbered from 0 to n-1, and an array edges where edges[i] = [fromi, toi] represents a directed edge from node fromi to node toi.

\n\n

Find the smallest set of vertices from which all nodes in the graph are reachable. It's guaranteed that a unique solution exists.

\n\n

Notice that you can return the vertices in any order.

\n\n

 

\n

Example 1:

\n\n

\"\"

\n\n
\nInput: n = 6, edges = [[0,1],[0,2],[2,5],[3,4],[4,2]]\nOutput: [0,3]\nExplanation: It's not possible to reach all the nodes from a single vertex. From 0 we can reach [0,1,2,5]. From 3 we can reach [3,4,2,5]. So we output [0,3].
\n\n

Example 2:

\n\n

\"\"

\n\n
\nInput: n = 5, edges = [[0,1],[2,1],[3,1],[1,4],[2,4]]\nOutput: [0,2,3]\nExplanation: Notice that vertices 0, 3 and 2 are not reachable from any other node, so we must include them. Also any of these vertices can reach nodes 1 and 4.\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "可以到达所有点的最少点数目", "translatedContent": "

给你一个 有向无环图 ， n 个节点编号为 0 到 n-1 ，以及一个边数组 edges ，其中 edges[i] = [fromi, toi] 表示一条从点  fromi 到点 toi 的有向边。

\n\n

找到最小的点集使得从这些点出发能到达图中所有点。题目保证解存在且唯一。

\n\n

你可以以任意顺序返回这些节点编号。

\n\n

 

\n\n

示例 1：

\n\n

\"\"

\n\n
输入：n = 6, edges = [[0,1],[0,2],[2,5],[3,4],[4,2]]\n输出：[0,3]\n解释：从单个节点出发无法到达所有节点。从 0 出发我们可以到达 [0,1,2,5] 。从 3 出发我们可以到达 [3,4,2,5] 。所以我们输出 [0,3] 。
\n\n

示例 2：

\n\n

\"\"

\n\n
输入：n = 5, edges = [[0,1],[2,1],[3,1],[1,4],[2,4]]\n输出：[0,2,3]\n解释：注意到节点 0，3 和 2 无法从其他节点到达，所以我们必须将它们包含在结果点集中，这些点都能到达节点 1 和 4 。\n
\n\n

 

\n\n

提示：

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